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Popular >

-cos(x)+4cos(2x)=0

Solution

- cos (x) + 4 cos (2 x) = 0

Solution

x = 0.68824 \dots + 2 πn, x = 2 π - 0.68824 \dots + 2 πn, x = 2.27491 \dots + 2 πn, x = - 2.27491 \dots + 2 πn

Degrees

x = 39.43339 \dots^{\circ} + 36 0^{\circ} n, x = 320.56660 \dots^{\circ} + 36 0^{\circ} n, x = 130.34311 \dots^{\circ} + 36 0^{\circ} n, x = - 130.34311 \dots^{\circ} + 36 0^{\circ} n

Solution steps

- cos (x) + 4 cos (2 x) = 0

Rewrite using trig identities

- cos (x) + 4 cos (2 x)

Use the Double Angle identity:

cos (2 x) = 2 cos^{2} (x) - 1

= - cos (x) + 4 (2 cos^{2} (x) - 1)

- cos (x) + (- 1 + 2 cos^{2} (x)) \cdot 4 = 0

Solve by substitution

- cos (x) + (- 1 + 2 cos^{2} (x)) \cdot 4 = 0

Let:

cos (x) = u

- u + (- 1 + 2 u^{2}) \cdot 4 = 0

- u + (- 1 + 2 u^{2}) \cdot 4 = 0 : u = \frac{1 + 129}{16}, u = \frac{1 - 129}{16}

- u + (- 1 + 2 u^{2}) \cdot 4 = 0

Expand

- u + (- 1 + 2 u^{2}) \cdot 4 : - u - 4 + 8 u^{2}

- u + (- 1 + 2 u^{2}) \cdot 4

= - u + 4 (- 1 + 2 u^{2})

Expand

4 (- 1 + 2 u^{2}) : - 4 + 8 u^{2}

4 (- 1 + 2 u^{2})

Apply the distributive law:

a (b + c) = ab + a c

a = 4, b = - 1, c = 2 u^{2}

= 4 (- 1) + 4 \cdot 2 u^{2}

Apply minus-plus rules

+ (- a) = - a

= - 4 \cdot 1 + 4 \cdot 2 u^{2}

Simplify

- 4 \cdot 1 + 4 \cdot 2 u^{2} : - 4 + 8 u^{2}

- 4 \cdot 1 + 4 \cdot 2 u^{2}

Multiply the numbers:

4 \cdot 1 = 4

= - 4 + 4 \cdot 2 u^{2}

Multiply the numbers:

4 \cdot 2 = 8

= - 4 + 8 u^{2}

= - 4 + 8 u^{2}

= - u - 4 + 8 u^{2}

- u - 4 + 8 u^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

8 u^{2} - u - 4 = 0

Solve with the quadratic formula

8 u^{2} - u - 4 = 0

Quadratic Equation Formula:

For

a = 8, b = - 1, c = - 4

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 8 ( - 4 )}{2 \cdot 8}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 8 ( - 4 )}{2 \cdot 8}

(- 1)^{2} - 4 \cdot 8 (- 4) = 129

(- 1)^{2} - 4 \cdot 8 (- 4)

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 8 \cdot 4

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 1)^{2} = 1^{2}

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 8 \cdot 4 = 128

4 \cdot 8 \cdot 4

Multiply the numbers:

4 \cdot 8 \cdot 4 = 128

= 128

= 1 + 128

Add the numbers:

1 + 128 = 129

= 129

u_{1, 2} = \frac{- ( - 1 ) \pm 129}{2 \cdot 8}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 129}{2 \cdot 8}, u_{2} = \frac{- ( - 1 ) - 129}{2 \cdot 8}

u = \frac{- ( - 1 ) + 129}{2 \cdot 8} : \frac{1 + 129}{16}

\frac{- ( - 1 ) + 129}{2 \cdot 8}

Apply rule

- (- a) = a

= \frac{1 + 129}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{1 + 129}{16}

u = \frac{- ( - 1 ) - 129}{2 \cdot 8} : \frac{1 - 129}{16}

\frac{- ( - 1 ) - 129}{2 \cdot 8}

Apply rule

- (- a) = a

= \frac{1 - 129}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{1 - 129}{16}

The solutions to the quadratic equation are:

u = \frac{1 + 129}{16}, u = \frac{1 - 129}{16}

Substitute back

u = cos (x)

cos (x) = \frac{1 + 129}{16}, cos (x) = \frac{1 - 129}{16}

cos (x) = \frac{1 + 129}{16}, cos (x) = \frac{1 - 129}{16}

cos (x) = \frac{1 + 129}{16} : x = arccos (\frac{1 + 129}{16}) + 2 πn, x = 2 π - arccos (\frac{1 + 129}{16}) + 2 πn

cos (x) = \frac{1 + 129}{16}

Apply trig inverse properties

cos (x) = \frac{1 + 129}{16}

General solutions for

cos (x) = \frac{1 + 129}{16}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{1 + 129}{16}) + 2 πn, x = 2 π - arccos (\frac{1 + 129}{16}) + 2 πn

x = arccos (\frac{1 + 129}{16}) + 2 πn, x = 2 π - arccos (\frac{1 + 129}{16}) + 2 πn

cos (x) = \frac{1 - 129}{16} : x = arccos (\frac{1 - 129}{16}) + 2 πn, x = - arccos (\frac{1 - 129}{16}) + 2 πn

cos (x) = \frac{1 - 129}{16}

Apply trig inverse properties

cos (x) = \frac{1 - 129}{16}

General solutions for

cos (x) = \frac{1 - 129}{16}

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

x = arccos (\frac{1 - 129}{16}) + 2 πn, x = - arccos (\frac{1 - 129}{16}) + 2 πn

x = arccos (\frac{1 - 129}{16}) + 2 πn, x = - arccos (\frac{1 - 129}{16}) + 2 πn

Combine all the solutions

x = arccos (\frac{1 + 129}{16}) + 2 πn, x = 2 π - arccos (\frac{1 + 129}{16}) + 2 πn, x = arccos (\frac{1 - 129}{16}) + 2 πn, x = - arccos (\frac{1 - 129}{16}) + 2 πn

Show solutions in decimal form

x = 0.68824 \dots + 2 πn, x = 2 π - 0.68824 \dots + 2 πn, x = 2.27491 \dots + 2 πn, x = - 2.27491 \dots + 2 πn

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